Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct __timestamper {};
void make_op(vector<int>& x, vector<pair<int, int>>& ans, int i, int j) {
  x[i] ^= x[j];
  ans.push_back(make_pair(i, j));
}
void make_swap(vector<int>& x, vector<pair<int, int>>& ans, int i, int j) {
  if (i == j) return;
  make_op(x, ans, i, j);
  make_op(x, ans, j, i);
  make_op(x, ans, i, j);
}
int n;
int make_iter(vector<int>& x, vector<pair<int, int>>& ans, int pos) {
  int mid = max_element(x.begin() + pos, x.end()) - x.begin();
  make_swap(x, ans, mid, pos);
  if (x[pos] == 0) return -1;
  int bit = 31 - __builtin_clz(x[pos]);
  for (int j = 0; j < n; j++)
    if (j != pos && (x[j] & (1 << bit))) make_op(x, ans, j, pos);
  return bit;
}
void solve() {
  vector<int> xs(n), ys(n);
  for (int& x : xs) scanf("%d", &x);
  for (int& y : ys) scanf("%d", &y);
  vector<pair<int, int>> ans1, ans2;
  for (int i = 0; i < n; i++) {
    int b1 = make_iter(xs, ans1, i);
    int b2 = make_iter(ys, ans2, i);
    if (b1 == -1 && b2 == -1) break;
    if (b1 == b2) continue;
    if (b2 > b1) {
      printf("-1\n");
      return;
    }
    for (int j = 0; j < i; j++) {
      if (ys[j] & (1 << b1)) make_op(xs, ans1, j, i);
    }
    make_op(xs, ans1, i, i);
    i--;
  }
  if (xs != ys) {
    printf("-1\n");
    return;
  }
  ans1.insert(ans1.end(), ans2.rbegin(), ans2.rend());
  printf("%d\n", ans1.size());
  for (auto x : ans1) printf("%d %d\n", x.first + 1, x.second + 1);
}
int main() {
  while (scanf("%d", &n) == 1) {
    solve();
  }
  return 0;
}